Suppose A is contained in the unit square of R^2, and the projection of A on any line outside the unit square is not Lebesgue measurable in R. Does that imply that A is not Lebesgue measurable in the plane?

No, it doesn't. Take a nonmeasurable set $S$ on a line segment, and make this line segment to be a side of $A$. Then $S$ as a subset of the plane is measurable (it is of measure zero), while its projection on the line segment (i.e. itself) is not. EDIT: As Pietro Mayer says one needs a union of a few isometric copies of my $S$ to yield an example whose projection on every line is nonmeasurable. 


This is an answer to Gerald Edgar's question (a) (see comment just after main question). To produce a set such that all but two projections are nonmeasurable, take a subset X of [0,1] that's not measurable in any interval and put a copy of X into the segment from (0,0) to (1,0) and a copy of the complement of X into the segment from (1,0) to (1,1). If we project to the Y axis we get two points, so it's measurable. If we project to the X axis, we get the interval [0,1], so it's measurable. If we project in any other direction, then the images of the two segments are intervals of positive length that do not share their endpoints, so by the local nonmeasurability of X the resulting projection is nonmeasurable. My guess is that similar tricks can be used to do other things but I haven't thought about this. Edit: I also have a very silly answer to Gerald Edgar's question (b), which is that the existence of such an example is consistent with ZF. It is known that there can be nonmeasurable sets of cardinality less than the continuum (of course, this means that CH fails). Let $K$ be such a set and let $X=K\times\mathbb{R}$. Now let $Y$ be the union of uncountably many rotations of $K$, but not continuum many. Finally, let $Z$ be the complement of $Y$ in the plane. If one of the rotates of $K$ is through an angle of $\theta$, then the projection of $Z$ on to the line $L_\theta$ that makes an angle of $\theta$ with the xaxis misses all points in $K$ (or rather the obvious copy of $K$ along that line). It also contains each point not in $K$, since if P is such a point, then we have removed fewer than continuum many points from the line perpendicular to $L_\theta$ that goes through P. Therefore, this projection is just a copy of the complement of $K$ and so nonmeasurable. If $L$ is any line that is not perpendicular to one of the $L_\theta$ for which $\theta$ is one of the angles we chose, then again we have removed fewer than continuum many points from $L$. Therefore, if we project onto $L_\phi$ for some $\phi$ that is not one of our chosen angles, then we obtain the whole of $L_\phi$, which is of course measurable. So we have uncountably many measurable projections and uncountably many nonmeasurable ones. I don't know what happens if we ask for continuum many measurable projections and continuum many nonmeasurable projections ... 


The subset S is not as required. Its projection on the Xaxis is not measurable, but it certainly does satisfy it for every line segment outsides the unit squrae. For instance, its progection on a line parallel to the yaxis is a single point, which is measurable in R. 


Pietro Majers does give a full and nice answer. I would like to ask further, what happens if we add the demand that A is not a union of less than $2^\aleph_0$many lines. Does this assumption, plus the assumption that the projection of A on every line outside the unit square is not Lebesgue measurable, ensures that A is not measurable in the plane? 


Partial answere (anyway I guess no). Take $S\subset [0,1]$ non mesaurabile and let $A:=S\times ${0}$\cup${0}$\times S, \ B:=\mathbb{R}\times ${0}$\cup${0}$\times \mathbb{R}$, then $A\subset B\subset \mathbb{R}^2$ and $A$ and $B$ have measure 0. If the projection is bijective is a omeomorphism from $B$ to the line, then the image of this proiection is no measurable: we have that $A\subset B$ isnt measurable (on $B$ consider the linear lebesgua measure, if we suppose $A$ measurable the also $A\setminus (0\times \mathbb{R})=S\times 0$ is measurable, absurd) further a projection is continuous, then meaurable i.e. the inverse image of a measurble subset is still measurable. If a projection is a omeomorphism then establishes an identification (or bijection) between the measurable sets. THe some follow if the line is vertical or horizontal (the image of prjection is essentally $S$). Let now $f: B\to l $ a projection , no injective and $l$ no vertical or horizontal. then $f(S\times 0)$ and $f(0\times S)$ are a homothetic (dilatate or traslate) copies of $S$. Then the problem reduces to finding a nomeasurable subset $ S $ such that the union of two of its homothetic images is nomeasurable. 

